GO Classes 2024 | Weekly Quiz 8 | Conditional Probability | Question: 8

Answer 1

 

For each level in the tree diagram the Sum of all  Node probability should be sum to 1 . 

0.15 + x + 0.45 + y = 1 ;

x+y = 0.4 

Comments

Very nice

Answer 2

$A$ and $B$ are independent boolean variables.

Using the $total$ $probability$,

$P(A) = P(A,B^{C}) + P(A,B) = 0.15 + 0.45 = 0.6$

$P(A^{C}) = P(A^{C},B) + P(A^{C},B^{C}) = x + y$

$\therefore x + y = P(A^{c}) = 1 – P(A) = 1- 0.6 = 0.4$

$Ans: 0.4$

Comments

Nice! This is how the question is meant to be solved.

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wow...I liked your observation.

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Thank you @Suraj Reddy & @Satyam Naik. 

It really means a lot.😊

Answer 3

solving first two equations we get P(B)=0.25 and P(A) = 0.6

x+y = $P(A^{C},B) + P(A^{{C}},B^{{C}})$

x+y = $P(A^{^{C^{}}})P(B)+P(A^{^{C^{}}})P(B^{^{C}})$

x+y = $P(A^{^{C^{}}})[P(B)+P(B^{^{C}})]$

x+y = $P(A^{C})$    

x+y = 0.4

Comments

how you calculated the p(b) plz tell

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divide equation 1 and 2 , P(A) gets canceled out you are left with $P(B)\div P(B^{C}) = 1/3$ now since $P(B^{C}) = 1 - P(B)$ substitute and cross multiply you will get P(B) = 0.25

P.S. – the question does not require us to find P(B)

Answer 4

P(A|B)=P(A∩B)/P(B),  P(A’|B)=1-P(A|B)=1-{P(A∩B)/P(B)}

 

 

 

          A

                        =>P(A’∩B)/P(B)={P(B)-P(A∩B)}/P(B) =>P(A’∩B)=P(B)-P(A∩B)...(i)
B                      P(A|B’)=P(A∩B’)/P(B’), P(A’|B’)=1-P(A|B’)=1-{P(A∩B’)/P(B’)}

                        =>P(A’∩B’)/P(B’)={P(B’)-P(A∩B’)}/P(B’) =>P(A’∩B’)=P(B’)-P(A∩B’)...(ii)
          A’           From (i)+(ii) we get that

 

 

          A           P(B)+P(B’)=P(A’∩B)+P(A∩B)+P(A’∩B’)+P(A∩B’)=P(A’,B)+P(A,B)+P(A’,B’)+P(A,B’)=x+0.15+y+0.45=x+y+0.6

B’                    1=x+y+0.6 =>x+y=0.4

          A’